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The following problem(p.668, 7) is from Integrals and Series [ Интегралы и ряды, А.П. Прудников, Ю.А. Брычков, О.И. Маричев.] states that

$$\sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}=\frac{\pi}{3}$$

How one can show that?

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Hint: re-write as $\frac{1}{3}\int_0^1 \sum x^k dx-2\int_0^1 \sum x^{2k}dx+\frac{8}{3} \int_0^1 \sum x^{4k}\,dx$. Can you see what to do next? –  L. F. Dec 10 '13 at 12:05
    
See the answers on this related post. –  Lucian Dec 10 '13 at 12:26
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I'd say the title would be more commonly translated as "Integrals and Series" –  DonAntonio Dec 10 '13 at 13:55
    
@DonAntonio: You are right! :) –  Salech Alhasov Dec 10 '13 at 13:57
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1 Answer 1

\begin{align} \sum_{k=0}^{\infty}\frac{1}{(k+1)(2k+1)(4k+1)}&=-\sum_{k=0}^\infty \frac{1}{k+1/2}+\frac{2}{3}\sum_{k=0}^{\infty} \frac{1}{k+1/4}+\frac{1}{3} \sum_{k=0}^\infty \frac{1}{k+1} \\ &=\frac{1}{3}\sum_{k=0}^\infty \frac{1}{k+1}-\frac{1}{k+1/2}+\frac{2}{3} \sum_{k=0}^{\infty} \frac{1}{k+1/4}-\frac{1}{k+1/2} \\ &=-\frac{1}{6}\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+1/2)}+\frac{1}{6}\sum_{k=0}^\infty \frac{1}{(k+1/4)(k+1/2)} \\ \text{Using Gauss's Digamma Theorem, }\\ &=-\frac{1}{6}\cdot \frac{\psi(1)-\psi(1/2)}{1/3-1/2}+ \frac{1}{6}\cdot \frac{\psi(1/4)-\psi(1/2)}{1/4-1/2} \\ &=-\frac{1}{6}\cdot \log 16+\frac{1}{6}\cdot 2(\pi+\log4)=\frac{\pi}{3} \end{align}

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